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Math Help - [Algebra?(rationalize)] I've never learnt this

  1. #1
    Junior Member Cthul's Avatar
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    [Algebra?(rationalize)] I've never learnt this


    I never been taught this and it's on the exam papers, how do I do this?
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  2. #2
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    Quote Originally Posted by Cthul View Post

    I never been taught this and it's on the exam papers, how do I do this?
    You'll need to get it of the form

    \frac{2(x^2 - 2x - 3) + R}{x^2 - 2x - 3} where R is some remainder...


    \frac{2x^2 + 2x + 1}{x^2 - 2x - 3} = \frac{2x^2 - 4x - 6 + 6x + 7}{x^2 - 2x - 3}

     = \frac{2(x^2 - 2x - 3) + 6x + 7}{x^2 - 2x - 3}

     = \frac{2(x^2 - 2x - 3)}{x^2 - 2x - 3} + \frac{6x + 7}{x^2 - 2x - 3}

     = 2 + \frac{6x + 7}{x^2 - 2x - 3}

     = 2 + \frac{6x + 7}{(x - 3)(x + 1)}.


    Now you need to break down the remainder using the method of partial fractions.

    Let \frac{A}{x - 3} + \frac{B}{x + 1} = \frac{6x + 7}{(x - 3)(x + 1)}

    \frac{A(x + 1) + B(x - 3)}{(x - 3)(x + 1)} = \frac{6x + 7}{(x - 3)(x + 1)}

    A(x + 1) + B(x - 3) = 6x + 7

    Ax + A + Bx - 3B = 6x + 7

    (A + B)x + A - 3B = 6x + 7.


    So

    A + B = 6

    A - 3B = 7.


    Subtracting the second equation from the first gives

    4B = -1

    B = -\frac{1}{4}.


    Since A + B = 6

    A - \frac{1}{4} = 6

    A = \frac{25}{4}.



    So finally we have:

    \frac{2x^2 + 2x + 1}{x^2 - 2x - 3} = 2 + \frac{6x + 7}{(x - 3)(x + 1)}

     = 2 + \frac{\frac{25}{4}}{(x - 3)} - \frac{\frac{1}{4}}{(x + 1)}

     = 2 + \frac{25}{4(x - 3)} - \frac{1}{4(x + 1)}.
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  3. #3
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    Quote Originally Posted by Cthul View Post

    I never been taught this and it's on the exam papers, how do I do this?
    First, divide x^2-2x-3 into 2x^2+2x+1

    You will get a remainder

    Hence \frac{2x^2+2x+1}{x^2-2x-3}=2+\frac{remainder}{(x-3)(x+1)}

    since x^2-2x-3=(x-3)(x+1)

    Now you need to express the remaining fraction as the sum of partial fractions

    \frac{remainder}{(x-3)(x+1)}=\frac{A}{x-3}+\frac{B}{x+1}

    Solve for A and B.
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  4. #4
    Junior Member Cthul's Avatar
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    Thanks

    This helped, thank you.
    (Solved)
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